The figure is not drawn to scale. EGHK is a parallelogram, DEK and EFK are isosceles triangles. KH//LT and ∠HST= 102°. DG is straight line.
- Find ∠DKE.
- Find ∠GFK.
(a)
∠HST = 102°
∠KPR = 102° (Corresponding angles)
∠NPK
= 180° - 102°
= 78° (Angles on a straight line)
∠DKE
= 180° - 78° - 78°
= 24° (Isosceles triangle)
(b)
∠NPK = ∠DEK = 78° (Corresponding angles)
∠KEF
= 180° - 78°
= 102° (Angles on a straight line)
∠EFR
= (180° - 102°) ÷ 2
= 39° (Isosceles triangle)
∠GFK
= 180° - 39°
= 141° (Angles on a straight line)
Answer(s): (a) 24°; (b) 141°